2025년 07월 미적분 30번

30. 함수 $\begin{array}{r}f(x)={\int }_0^x{e}^{\cos \pi t}dt\end{array}$$\begin{aligned} f(x)=\int_0^x e^{\cos\pi t}dt\end{aligned}$의 역함수를 $g(x)$$g(x)$라 할 때, 실수 전체의 집합에서 도함수가 연속인 함수 $h(x)$$h(x)$가 모든 실수 $x$$x$에 대하여 $h(g(x)+2)=2x^3+6f(1)x^2+1$$h(g(x)+2)=2x^3 +6f(1)x^2 +1$을 만족시킨다. $\begin{array}{r}{\int }_3^7\frac{h^{\prime }(x)}{f(x)}dx=k\times \{f(1){\}}^2\end{array}$$\begin{aligned} \int_3^7 \cfrac {h'(x)}{f(x)}dx =k\times \{f(1)\}^2 \end{aligned}$일 때, 실수 $k$$k$의 값을 구하시오.

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i. 정리

• $f(x)$$f(x)$에 대해 살펴보자.

  • $f(0)=0$$f(0)=0$

  • $f^{\prime }(x)=e^{\cos \pi x}$$f'(x)=e^{\cos \pi x}$

  • $f^{\prime }(x+2)=f^{\prime }(x)$$f'(x+2)=f'(x)$ 주기함수인가?

    $f(x+2)=f(x)+C$$f(x+2)=f(x)+C$

    $f(2)=f(0)+C$$f(2)=f(0)+C$

    $\begin{array}{r}f(2)={\int }_0^2={\int }_0^1+{\int }_1^2={\int }_0^1+{\int }_{-1}^0\end{array}$$\begin{aligned} f(2)=\int_0^2 =\int_0^1 +\int_1^2 =\int_0^1+\int_{-1}^0\end{aligned}$

  • $f^{\prime }(-x)=f(x)$$f'(-x)=f(x)$ $y$$y$축 대칭이네?

    어? $\begin{array}{r}f(2)={\int }_0^1+{\int }_{-1}^0={\int }_{-1}^1=2{\int }_0^1\end{array}$$\begin{aligned}f(2)=\int_0^1+\int_{-1}^0 =\int_{-1}^1=2\int_0^1\end{aligned}$

    즉, $f(2)=2f(1)$$f(2)=2f(1)$

    쓸모없나?

• $g(f(x))=x,f(g(x))=x$$g(f(x))=x,~f(g(x))=x$

• $h(g(x)+2)=2x^3+6f(1)x^2+1$$h(g(x)+2)=2x^3 +6f(1)x^2 +1$에서 $x$$x$대신 $f(x)$$f(x)$를 대입하면,

  $\begin{array}{rl}h(g(f(x))+2)& =2f(x{)}^3+6f(1)f(x{)}^2+1\\ & =h(x+2)\end{array}$$\begin{aligned}h(g(f(x))+2)&=2f(x)^3 +6f(1)f(x)^2 +1\\ &=h(x+2)\end{aligned}$

ii. 접근

• $h^{\prime }(x)$$h'(x)$를 구하기 위해서 미분하자.

  $\begin{array}{rl}{h}^{\prime }(x+2)& =6f(x{)}^2{f}^{\prime }(x)+12f(1)f(x)f^{\prime }(x)\\ & =6f(x)f^{\prime }(x)\{f(x)+2f(1)\}\end{array}$$\begin{aligned} h'(x+2)&=6f(x)^2f'(x)+12f(1)f(x)f'(x) \\ &= 6f(x)f'(x)\{f(x)+2f(1)\big\}\end{aligned}$

  어? 뭔가 될 거 같다?

• $x=t+2$$x=t+2$로 치환하자.

  $dx=dt$$dx=dt$이고 적분 구간은 $1$$1$에서 $5$$5$

  $\begin{array}{rl}{\int }_3^7\frac{h^{\prime }(x)}{f(x)}dx& ={\int }_1^5\frac{h^{\prime }(x+2)}{f(x+2)}\end{array}$$\begin{aligned}\int_3^7 \cfrac{h'(x)}{f(x)} dx &=\int_1^5\cfrac {h'(x+2)}{f(x+2)} \end{aligned}$

• $f(x+2)$$f(x+2)$를 ....

  $f(x+2)=f(x)+C$$f(x+2)=f(x)+C$에서

  $f(2)=f(0)+C=C$$f(2)=f(0)+C=C$이고 $f(2)=2f(1)$$f(2)=2f(1)$

• 다시 식을 정리하면,

  $\begin{array}{rl}\text{준식}& ={\int }_1^{5}6f(x)f^{\prime }(x)dx\\ & =[3f(x{)}^2{]}_1^5\\ & =3(f(5{)}^2-f(1{)}^2)\end{array}$$\begin{aligned}\text{준식}&=\int_1^5 6f(x)f'(x)dx\\ &= \Big[3f(x)^2 \Big]_1^5\\ &= 3(f(5)^2 -f(1)^2) \end{aligned}$

• $f(5)$$f(5)$를 구하자...

  $\begin{array}{rl}f(5)={\int }_0^5& ={\int }_0^1+{\int }_1^2+{\int }_2^3+{\int }_3^4+{\int }_4^5\\ & ={\int }_0^1+{\int }_{-1}^0+{\int }_0^1+\cdots \\ & =5{\int }_0^1\end{array}$$\begin{aligned}f(5)=\int_0^5&=\int_0^1+\int_1^2+\int_2^3+\int_3^4+\int_4^5\\ &= \int_0^1 +\int_{-1}^0+\int_0^1+\cdots \\ &= 5\int_0^1\end{aligned}$

  $\therefore f(5)=5f(1)$$\therefore~ f(5)=5f(1)$

• 마지막 계산

  $\begin{array}{rl}\text{준식}& =3((5f(1){)}^2-f(1{)}^2)\\ & =72f(1{)}^2\end{array}$$\begin{aligned}\text{준식}&=3((5f(1))^2 -f(1)^2)\\ &= 72 f(1)^2 \end{aligned}$

$\therefore k=72$$\therefore~ k=72$